A chemist is making a solution that is to be 55% chlorine. He has one solution that is 40% chlorine and another that is 65% chlorine. He wants 100 liters of the final solution. How much of each should he mix?

Aug 11, 2015

You need to mix 40 L of 40% solution and 60 L of 65% solution.

Explanation:

You know that the volume of the final solution is 100 L. If $x$ represents the volume of the 40% solution and $y$ represents the volume of the 65% solution, then you can say that

$x + y = 100$

Now you need to focus on the chlorine content. Since the first solution is 40% chlorine, then for $x$ liters you would get

$\left(x\right) \text{cancel("L solution") * "40 g chlorine"/(100cancel("L solution")) = (0.4 * x)" g chlorine}$

The same can be said for the 65% solution

$\left(y\right) \text{cancel("L solution") * "65 g chlorine"/(100cancel("L solution")) = (0.65 * y)" g chlorine}$

The final solution has a percent concentration of 55% chlorine, which means that it must contain

$100 \text{cancel("L solution") * "55 g chlorine"/(100cancel("L solution")) = "55 g chlorine}$

The second equation you can write will thus relate the amount of chlorine each solution contributes to the final solution

$0.4 x + 0.65 y = 55$

This is your system of equations

$\left\{\begin{matrix}x + y = 100 \\ 0.4 x + 0.65 y = 55\end{matrix}\right.$

Use the first one to write $x$ as a function of $y$

$x = 100 - y$

Now use this expression in the second equation to get the value of $y$

$0.4 \cdot \left(100 - y\right) + 0.65 y = 55$

$40 - 0.4 y + 0.65 y = 55$

$0.25 y = 15 \implies y = \frac{15}{0.25} = \textcolor{g r e e n}{60}$

This means that $x$ is equal to

$x = 100 - 60 = \textcolor{g r e e n}{40}$

So, if you mix 40 L of a 40% chlorine solution with 60 L of a 65% chlorine solution you will get 100 L of a 55% chlorine solution.